Questions: In the laboratory a "coffee cup" calorimeter, or constant pressure calorimeter, is frequently used to determine the specific heat of a solid, or to measure the energy of a solution phase reaction. A chunk of iron weighing 18.61 grams and originally at 97.20°C is dropped into an insulated cup containing 83.84 grams of water at 21.11°C. The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.67 J/°C. Using the accepted value for the specific heat of iron (See the References tool), calculate the final temperature of the water. Assume that no heat is lost to the surroundings.

Solution Steps

• Mass of iron (m_iron) = 18.61 grams
• Initial temperature of iron (T_initial_iron) = 97.20 °C
• Mass of water (m_water) = 83.84 grams
• Initial temperature of water (T_initial_water) = 21.11 °C
• Heat capacity of the calorimeter (C_calorimeter) = 1.67 J/°C
• Specific heat of iron (c_iron) = 0.449 J/g°C
• Specific heat of water (c_water) = 4.18 J/g°C

The heat lost by the iron will be gained by the water and the calorimeter: \[ q_{iron} = - (q_{water} + q_{calorimeter}) \]

\[ q_{iron} = m_{iron} \cdot c_{iron} \cdot (T_{final} - T_{initial_iron}) \] \[ q_{water} = m_{water} \cdot c_{water} \cdot (T_{final} - T_{initial_water}) \] \[ q_{calorimeter} = C_{calorimeter} \cdot (T_{final} - T_{initial_water}) \]

\[ m_{iron} \cdot c_{iron} \cdot (T_{final} - T_{initial_iron}) = - (m_{water} \cdot c_{water} \cdot (T_{final} - T_{initial_water}) + C_{calorimeter} \cdot (T_{final} - T_{initial_water})) \]

\[ 18.61 \cdot 0.449 \cdot (T_{final} - 97.20) = - (83.84 \cdot 4.18 \cdot (T_{final} - 21.11) + 1.67 \cdot (T_{final} - 21.11)) \]

\[ 8.35389 \cdot (T_{final} - 97.20) = - (350.4512 \cdot (T_{final} - 21.11) + 1.67 \cdot (T_{final} - 21.11)) \] \[ 8.35389 \cdot T_{final} - 812.84 = - (352.1212 \cdot T_{final} - 7427.76) \] \[ 8.35389 \cdot T_{final} - 812.84 = - 352.1212 \cdot T_{final} + 7427.76 \] \[ 8.35389 \cdot T_{final} + 352.1212 \cdot T_{final} = 7427.76 + 812.84 \] \[ 360.47509 \cdot T_{final} = 8240.60 \] \[ T_{final} = \frac{8240.60}{360.47509} \] \[ T_{final} \approx 22.86 \, °C \]

Final Answer